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OCR MEI Further Numerical Methods 2020 November Q6
10 marks Standard +0.3
6 Fig. 6.1 shows the graph of \(y = \mathrm { e } ^ { 3 x } - 11 x - 0.5\) for \(- 0.5 \leqslant x \leqslant 1\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{87bb8eb7-b725-48b0-b32b-0bfce624cd91-08_576_881_315_333} \captionsetup{labelformat=empty} \caption{Fig. 6.1}
\end{figure} The equation \(\mathrm { e } ^ { 3 x } - 11 x - 0.5 = 0\) has two roots, \(\alpha\) and \(\beta\), such that \(\alpha < \beta\). Dennis is going to use the method of interval bisection with starting values denoted by \(a\) and \(b\).
  1. Explain why the method of interval bisection starting with \(a = 0\) and \(b = 1\) may not be used to find either \(\alpha\) or \(\beta\). Dennis uses the method of interval bisection starting with \(a = 0\) and \(b = 0.5\) to find \(\alpha\). Some spreadsheet output is shown in Fig. 6.2. \begin{table}[h]
    ABCDEF
    1af(a)bf(b)\(x _ { \text {new } }\)\(\mathrm { f } \left( x _ { \text {new } } \right)\)
    200.50.5-1.518310.25-1.133
    300.50.25-1.1330.125-0.42
    400.50.125-0.420010.06250.01873
    50.06250.018730.125-0.420010.09375-0.2065
    \captionsetup{labelformat=empty} \caption{Fig. 6.2}
    \end{table} Dennis states that the formula in cell B2 is $$= \operatorname { EXP } \left( 3 ^ { * } \mathrm {~A} 1 \right) - 11 \mathrm {~A} 2 - 0.5$$ Dennis has made two errors.
  2. Write a correct version of Dennis's formula for cell B2. The formula in cell A3, which is correct, is
    = IF(F2 > 0, E2, A2)
  3. Write a suitable formula for cell C3.
  4. Use the information in Fig. 6.2 to
    Liren uses a different method to find a sequence of estimates of the value of \(\beta\) using a spreadsheet. The output, together with some further analysis, is shown in Fig. 6.3. \begin{table}[h]
    ABCD
    1\(x\)f(x)differenceratio
    20.4-1.5799
    30.6-1.0504
    40.996718.4245
    50.64398-0.6809-0.35273
    60.67036-0.40260.026378-0.0748
    70.708520.083860.0381641.44682
    80.70194-0.0075-0.00658-0.1724
    90.70248-0.00010.00054-0.082
    100.70249\(1.8 \mathrm { E } - 07\)\(8.88 \mathrm { E } - 06\)0.01646
    \captionsetup{labelformat=empty} \caption{Fig. 6.3}
    \end{table} The formula in cell A4 is $$= ( \mathrm { A } 2 * \mathrm {~B} 3 - \mathrm { A } 3 * \mathrm {~B} 2 ) / ( \mathrm { B } 3 - \mathrm { B } 2 )$$
  5. State the method being used.
  6. Explain what the values in column D tell you about the order of convergence of this sequence of estimates. Liren states that \(\beta = 0.70249\) correct to 5 decimal places.
  7. Determine whether Liren is correct.
OCR MEI Further Numerical Methods 2020 November Q7
11 marks Challenging +1.2
7 Fig. 7.1 shows two values of \(x\) and the associated values of \(\mathrm { f } ( x )\). \begin{table}[h]
\(x\)33.5
\(\mathrm { f } ( x )\)6.0827634.596194
\captionsetup{labelformat=empty} \caption{Fig. 7.1}
\end{table}
  1. Use the forward difference method to calculate an estimate of the gradient of \(\mathrm { f } ( x )\) at \(x = 3\), giving your answer correct to 4 decimal places. Fig. 7.2 shows some spreadsheet output with additional values of \(x\) and the associated values of \(\mathrm { f } ( x )\). \begin{table}[h]
    \(x\)33.000013.00013.0013.013.1
    \(\mathrm { f } ( x )\)6.0827636.082746.0825416.080546.0604545.848846
    \captionsetup{labelformat=empty} \caption{Fig. 7.2}
    \end{table} These values have been used to produce a sequence of estimates of the gradient of \(\mathrm { f } ( x )\) at \(x = 3\), together with some further analysis. This is shown in the spreadsheet output in Fig. 7.3. \begin{table}[h]
    \(h\)0.10.010.0010.00010.00001
    estimate-2.339165-2.230883-2.220532-2.219501-2.219398
    difference0.10828150.0103520.00103070.000103
    ratio0.0956020.0995670.0999568
    \captionsetup{labelformat=empty} \caption{Fig. 7.3}
    \end{table} Tommy states that the differences between successive estimates is decreasing so rapidly that the order of convergence of this sequence of estimates is much faster than first order.
  2. Explain whether or not Tommy is correct.
  3. Use extrapolation to determine the value of the gradient of \(\mathrm { f } ( x )\) at \(x = 3\) as accurately as possible, justifying the precision quoted.
  4. Calculate an estimate of the absolute error when \(\mathrm { f } ( 3 )\) is used as an approximation to \(\mathrm { f } ( 3.02 )\).
OCR MEI Further Numerical Methods 2021 November Q1
5 marks Standard +0.3
1
    1. Determine the relative error when
      \begin{table}[h]
      ABC
      121.4142142
      \captionsetup{labelformat=empty} \caption{Fig. 1}
      \end{table} The formula in cell B1 is = SQRT (A1)
      and the formula in cell C 1 is \(\quad = \mathrm { B } 1 \wedge 2\).
      Ben evaluates \(1.414214 ^ { 2 }\) on his calculator and obtains 2.000001238 . He states that this shows that the value displayed in cell C1 is wrong. Explain whether Ben is correct.
OCR MEI Further Numerical Methods 2021 November Q2
6 marks Challenging +1.2
2 The table shows some values of \(x\) and the associated values of \(\mathrm { f } ( x )\).
\(x\)12345
\(\mathrm { f } ( x )\)- 0.65- 0.351.775.7111.47
  1. Complete the difference table in the Printed Answer Booklet.
  2. Explain why the data may be interpolated by a polynomial of degree 2.
  3. Use Newton's forward difference interpolation formula to obtain a polynomial of degree 2 for the data.
OCR MEI Further Numerical Methods 2021 November Q3
7 marks Standard +0.3
3 The method of False Position is used to find a sequence of approximations to the root of an equation. The spreadsheet output showing these approximations, together with some further analysis, is shown below.
CDEFGHIJ
4af(a)b\(\mathrm { f } ( b )\)\(x _ { \text {new } }\)\(\mathrm { f } \left( x _ { \text {new } } \right)\)differenceratio
51-1.8248217.28991.09547-1.80507
61.09547-1.80507217.28991.18097-1.754180.08551
71.18097-1.75418217.28991.25641-1.662460.075440.88229
81.25641-1.66246217.28991.32164-1.527810.065230.86458
91.32164-1.52781217.28991.37672-1.357060.055080.84439
101.37672-1.35706217.28991.42208-1.16420.045360.8236
111.42208-1.1642217.28991.45853-0.966160.036460.80376
121.45853-0.96616217.28991.48719-0.778250.028660.78598
131.48719-0.77825217.2899
14
The formula in cell D5 is \(\quad = \mathrm { SINH } \left( \mathrm { C5 } ^ { \wedge } 2 \right) - \mathrm { C5 } ^ { \wedge } 3 - 2\).
  1. Write down the equation which is being solved. The formula in cell C 6 is \(\quad = \mathrm { IF } ( \mathrm { H } 5 < 0 , \mathrm { G } 5 , \mathrm { C } 5 )\).
  2. Write down a similar formula for cell E6.
  3. Calculate the values which would be displayed in cells G13 and G14 to find further approximations to the root.
  4. Explain what the values in column J tell you about
OCR MEI Further Numerical Methods 2021 November Q4
6 marks Standard +0.3
4 The table shows some values of \(x\) and the associated values of \(\mathrm { f } ( x )\).
\(x\)44.00014.0014.014.1
\(\mathrm { f } ( x )\)44.00023864.00238714.02394684.2472072
  1. Calculate four estimates of the derivative of \(\mathrm { f } ( x )\) at \(x = 4\).
  2. Without doing any further calculation, state the value of \(f ^ { \prime } ( 4 )\) as accurately as you can, justifying the precision quoted.
OCR MEI Further Numerical Methods 2021 November Q5
9 marks Standard +0.3
5 When Nina does the weekly grocery shopping she models the total cost by adding up the cost of each item in her head as she goes along. To simplify matters she rounds the cost of each item to the nearest pound. One week Nina buys 48 items.
  1. Calculate the maximum possible error in Nina's model in this case. Nina estimated the total cost of her shopping to be \(\pounds 92\). The actual cost is \(\pounds 90.23\).
  2. Explain whether this is consistent with Nina's model. The next week her husband, Kareem, does the weekly shopping. He models the total cost by chopping the cost of each item to the nearest pound as he goes along. On this occasion Kareem buys 52 items.
  3. Calculate the expected error in Kareem's model in this case. Using his model Kareem estimates the total cost as \(\pounds 76\). The total cost of the shopping is \(\pounds 103.24\).
  4. Explain how such a large error could arise. The next week Kareem buys \(n\) items.
  5. Write down a formula for the maximum possible error when Kareem uses his model to estimate the total cost of his shopping.
  6. Explain how Kareem's model could be adapted so that his formula gives the same expected error as Nina's model when they are both used to estimate the total cost of the shopping.
OCR MEI Further Numerical Methods 2021 November Q6
12 marks Challenging +1.2
6 The equation \(0.5 \ln x - x ^ { 2 } + x + 1 = 0\) has two roots \(\alpha\) and \(\beta\), such that \(0 < \alpha < 1\) and \(1 < \beta < 2\).
  1. Use the Newton-Raphson method with \(x _ { 0 } = 1\) to obtain \(\beta\) correct to \(\mathbf { 6 }\) decimal places. Fig. 6.1 shows part of the graph of \(y = 0.5 \ln x - x ^ { 2 } + x + 1\). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{945883ad-c153-4c51-83d3-978e4c769ed5-06_1112_1156_529_354} \captionsetup{labelformat=empty} \caption{Fig. 6.1}
    \end{figure}
  2. On the copy of Fig. 6.1 in the Printed Answer Booklet, illustrate the Newton-Raphson method working to obtain \(x _ { 1 }\) from \(x _ { 0 } = 1\). Beth is trying to find \(\alpha\) correct to 6 decimal places.
  3. Suggest a reason why she might choose the Newton-Raphson method instead of fixed point iteration. Beth tries to find \(\alpha\) using the Newton-Raphson method with a starting value of \(x _ { 0 } = 0.5\). Her spreadsheet output is shown in Fig. 6.2. \begin{table}[h]
    \(r\)\(\mathrm { x } _ { \mathrm { r } }\)
    00.5
    1- 0.40343
    2\#NUM!
    \captionsetup{labelformat=empty} \caption{Fig. 6.2}
    \end{table}
  4. Explain how the display \#NUM! has arisen in the cell for \(x _ { 2 }\). Beth decides to use the iterative formula $$x _ { n + 1 } = g \left( x _ { n } \right) = \sqrt { 0.5 \ln \left( x _ { n } \right) + x _ { n } + 1 }$$
  5. Determine the outcome when Beth uses this formula with \(x _ { 0 } = 0.5\).
  6. Use the relaxed iteration \(\mathrm { x } _ { \mathrm { n } + 1 } = ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } } + \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right)\) with \(\lambda = - 0.041\) and \(x _ { 0 } = 0.5\) to obtain \(\alpha\) correct to \(\mathbf { 6 }\) decimal places.
OCR MEI Further Numerical Methods 2021 November Q7
15 marks Standard +0.8
7 Sarah uses the trapezium rule to find a sequence of approximations to \(\int _ { 0 } ^ { 1 } \sqrt { \tanh ( x ) } \mathrm { d } x\).
Her spreadsheet output is shown in Fig. 7.1. \begin{table}[h]
\(n\)\(T _ { n }\)differenceratio
10.43634681
20.55806940.121723
40.601998430.0439290.36089
80.617870730.0158720.36132
160.623576010.0057050.35945
320.625617160.0020410.35777
\captionsetup{labelformat=empty} \caption{Fig. 7.1}
\end{table}
  1. Write down the value of \(h\) used to find the approximation 0.62357601 .
  2. Without doing any further calculation, state the value of \(\int _ { 0 } ^ { 1 } \sqrt { \tanh ( x ) } \mathrm { d } x\) as accurately as you
    can, justifying the precision quoted.
  3. Explain what the values in the ratio column tell you about the order of convergence of this sequence of approximations. Sarah carries out further work using the midpoint rule and Simpson's rule. Her results are shown in Fig. 7.2. \begin{table}[h]
    MNOPQR
    1\(n\)\(T _ { n }\)\(M _ { n }\)\(S _ { 2 n }\)differenceratio
    210.436346810.6797920.5986436
    320.55806940.645927450.616641440.018
    440.601998430.633743040.62316150.006520.362269
    580.617870730.629281290.625477770.002320.355253
    6160.623576010.627658310.626297550.000820.35392
    7320.625617160.62707259
    \captionsetup{labelformat=empty} \caption{Fig. 7.2}
    \end{table}
  4. Write down an efficient spreadsheet formula for calculating \(S _ { 16 }\).
  5. Determine the missing values in row 7.
  6. Use extrapolation to determine the value of \(\int _ { 0 } ^ { 1 } \sqrt { \tanh ( x ) } d x\) as accurately as you can, justifying
    the precision quoted.
    [0pt] [6]
OCR MEI Further Numerical Methods Specimen Q1
5 marks Standard +0.8
1
  1. Solve the following simultaneous equations. $$\begin{aligned} & x + \quad y = 1 \\ & x + 0.99 y = 2 \end{aligned}$$
  2. The coefficient 0.99 is correct to two decimal places. All other coefficients in the equations are exact. With the aid of suitable calculations, explain why your answer to part (i) is unreliable.
OCR MEI Further Numerical Methods Specimen Q2
4 marks Moderate -0.3
2 The following spreadsheet printout shows the bisection method being applied to the equation \(\mathrm { f } ( x ) = 0\), where \(\mathrm { f } ( x ) = \mathrm { e } ^ { x } - x ^ { 2 } - 2\). Some values of \(\mathrm { f } ( x )\) are shown in columns B and D.
ABCDEFG
1a\(\mathrm { f } ( a )\)bf(b)\(( a + b ) / 2\)\(\mathrm { f } ( ( a + b ) / 2 )\)mpe
21-0.2817221.3890561.50.2316890.5
31-0.281721.50.2316891.25-0.0721570.25
41.25-0.072161.50.2316891.3750.0644520.125
51.25-0.072161.3750.0644521.3125-0.0072060.0625
61.3125-0.007211.3750.0644521.343750.0277280.03125
  1. The formula in cell A 3 is \(= \mathrm { IF } ( \mathrm { F } 2 > 0\), A2, E2). State the purpose of this formula.
  2. The formula in cell C 3 is \(= \mathrm { IF } ( \mathrm { F } 2 > 0 , \ldots , \ldots )\). What are the missing cell references?
  3. In which row is the magnitude of the maximum possible error (mpe) less than \(5 \times 10 ^ { - 7 }\) for the first time?
OCR MEI Further Numerical Methods Specimen Q3
4 marks Challenging +1.2
3 The equation \(\sinh x + x ^ { 2 } - 1 = 0\) has a root, \(\alpha\), such that \(0 < \alpha < 1\).
  1. Verify that the iteration \(x _ { r + 1 } = \frac { 1 - \sinh x _ { r } } { x _ { r } }\) with \(x _ { 0 } = 1\) fails to converge to this root.
  2. Use the relaxed iteration \(x _ { r + 1 } = ( 1 - \lambda ) x _ { r } + \lambda \left( \frac { 1 - \sinh x _ { r } } { x _ { r } } \right)\) with \(\lambda = \frac { 1 } { 4 }\) and \(x _ { 0 } = 1\) to find \(\alpha\) correct to 6 decimal places.
OCR MEI Further Numerical Methods Specimen Q4
6 marks Standard +0.8
4 The table below gives values of a function \(y = \mathrm { f } ( x )\).
\(x\)0.20.30.350.40.450.50.6
\(\mathrm { f } ( x )\)0.7899220.7546280.7491990.7499970.7562570.7675230.804299
  1. Calculate three estimates of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at \(x = 0.4\) using the central difference method.
  2. State the value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) at \(x = 0.4\) to an appropriate degree of accuracy. Justify your answer.
OCR MEI Further Numerical Methods Specimen Q5
10 marks Standard +0.3
5 A vehicle is moving in a straight line. Its velocity at different times is recorded and shown below. The velocities are recorded to 5 significant figures and the times may be assumed to be exact.
Time \(( t\) seconds \()\)5101215
Velocity \(( v\) metres per second \()\)5.125011.00014.00018.375
It is suggested initially that a quadratic model may be appropriate for this situation.
  1. Given that the vehicle is modelled as a particle with constant mass, what assumption about the net force acting on the vehicle leads to a quadratic model?
  2. Find Newton's interpolating polynomial of degree 2 to model this situation. Write your answer in the form \(v = a t ^ { 2 } + b t + c\).
  3. Comment on whether this model appears to be appropriate.
  4. Use this model to find an approximation to the distance travelled over the interval \(5 \leq t \leq 15\). Further investigation suggests that a cubic model may be more appropriate.
  5. What technique would you use to fit a cubic model to the data in the table?
OCR MEI Further Numerical Methods Specimen Q6
12 marks Standard +0.8
6 The secant method is to be used to solve the equation \(x - \ln ( \cos x ) - 1 = 0\).
  1. Show that starting with \(x _ { 0 } = - 1\) and \(x _ { 1 } = 0\) leads to the method failing to find the root between \(x = 0\) and \(x = 1\). The spreadsheet printout shows the application of the secant method starting with \(x _ { 0 } = 0\) and \(x _ { 1 } = 1\). Successive approximations to the root are in column E.
    ABCDE
    1\(x _ { n }\)\(\mathrm { f } \left( x _ { n } \right)\)\(x _ { n + 1 }\)\(\mathrm { f } \left( x _ { n + 1 } \right)\)\(x _ { n + 2 }\)
    20-110.61562650.6189549
    310.61562650.6189549-0.1758460.7036139
    40.6189549-0.17584610.7036139-0.0252450.7178053
    50.7036139-0.02524510.71780530.00116190.7171808
    60.71780530.00116190.7171808- 7.4 E -060.7171848
    70.7171808-7.402E-060.7171848-2.16E-090.7171848
    80.7171848-2.16E-090.71718483.997 E -150.7171848
  2. What feature of column B shows that this application of the secant method has been successful?
  3. Write down a suitable spreadsheet formula to obtain the value in cell E2. Some analysis of convergence is carried out, and the following spreadsheet output is obtained.
    ABCDEFGH
    1\(x _ { n }\)\(\mathrm { f } \left( x _ { n } \right)\)\(x _ { n + 1 }\)\(\mathrm { f } \left( x _ { n + 1 } \right)\)\(x _ { n + 2 }\)
    20-110.61562650.61895490.0846590.1676291.980053
    310.61562650.6189549-0.1758460.70361390.0141913-0.044-3.10054
    40.6189549-0.17584610.7036139-0.0252450.7178053-0.0006244-0.0063310.13727
    50.7036139-0.02524510.71780530.00116190.71718083.953 E -060.00029273.83899
    60.71780530.00116190.7171808- 7.4 E -060.71718481.154 E -09-1.8E-06
    70.7171808-7.402E-060.7171848- 2.16 E -090.7171848- 2.109 E -15
    80.7171848- 2.16 E -090.71718483.997 E -150.7171848
    The formula in cell F2 is =E3-E2. The formula in cell G2 is =F3/F2. The formula in cell H2 is =F3/(F2\^{}2).
  4. (A) Explain the purpose of each of these three formulae.
    (B) Explain the significance of the values in columns G and H in terms of the rate of convergence of the secant method.
  5. Explain why the values in cells F6 and F7 are not 0 . [Question 7 is printed overleaf.]
OCR MEI Further Numerical Methods Specimen Q7
19 marks Challenging +1.2
7 Fig. 7 shows the graph of \(y = \mathrm { f } ( x )\) for values of \(x\) from 0 to 1 . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{662c2d48-228a-4b94-a4b4-cdd31634ef21-6_693_673_390_696} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure} The following spreadsheet printout shows estimates of \(\int _ { 0 } ^ { 1 } f ( x ) d x\) found using the midpoint and trapezium rules for different values of \(h\), the strip width.
AABC
1\(h\)MidpointTrapezium
211.998517421.751283839
30.51.96385911.874900631
40.251.951352591.919379864
50.1251.946821021.935366229
  1. Without doing any further calculation, write down the smallest possible interval which contains the value of the integral. Justify your answer.
  2. (A) - Calculate the ratio of differences, \(r\), for the sequence of estimates calculated using the trapezium rule.
    Using a similar approach with the sequence of estimates calculated using the midpoint rule, the approximation to the integral from extrapolation was found to be 1.94427 correct to 5 decimal places.
  3. Andrea uses the extrapolated midpoint rule value and the value found in part (ii) ( \(B\) ) to write down an interval which contains the value of the integral. Comment on the validity of Andrea's method.
  4. Use the values from the spreadsheet output to calculate an approximation to the integral using Simpson's rule with \(h = 0.125\). Give your answer to 5 decimal places. Approximations to the integral using Simpson's rule are given in the spreadsheet output below. The number of applications of Simpson's rule is given in column N.
    NOPQ
    \(n\)Simpsondifferencesratio
    11.916106230.018100050.3584931
    21.934206280.006488740.3556525
    41.940695020.002307740.3544828
    81.943002750.000818050.3539885
    161.943820810.000289580.3537638
    321.944110390.000102440.3536568
    641.94421283\(3.623 \mathrm { E } - 05\)
    1281.94424906
  5. Use the spreadsheet output to find the value of the integral as accurately as possible. Justify the precision quoted. \section*{END OF QUESTION PAPER} }{www.ocr.org.uk}) after the live examination series. If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.
    OCR is part of the }\section*{}
OCR MEI Further Extra Pure 2022 June Q1
7 marks Standard +0.3
1 Three sequences, \(\mathrm { a } _ { \mathrm { n } } , \mathrm { b } _ { \mathrm { n } }\) and \(\mathrm { c } _ { \mathrm { n } }\), are defined for \(n \geqslant 1\) by the following recurrence relations. $$\begin{aligned} & \left( a _ { n + 1 } - 2 \right) \left( 2 - a _ { n } \right) = 3 \text { with } a _ { 1 } = 3 \\ & b _ { n + 1 } = - \frac { 1 } { 2 } b _ { n } + 3 \text { with } b _ { 1 } = 1.5 \\ & c _ { n + 1 } - \frac { c _ { n } ^ { 2 } } { n } = 1 \text { with } c _ { 1 } = 2.5 \end{aligned}$$ The output from a spreadsheet which presents the first 10 terms of \(a _ { n } , b _ { n }\) and \(c _ { n }\), is shown below.
ABCD
1\(n\)\(a _ { n }\)\(b _ { n }\)\(c _ { n }\)
2131.52.5
32-12.257.25
4331.87527.28125
54-12.0625249.0889
6531.9687515512.32
76-12.0156348126390
8731.992193.86E+14
98-12.00391\(2.13 \mathrm { E } + 28\)
10931.998055.66E+55
1110-12.000983.6E+110
Without attempting to solve any recurrence relations, describe the apparent behaviour, including as \(n \rightarrow \infty\), of
  • \(a _ { n }\)
  • \(\mathrm { b } _ { \mathrm { n } }\)
  • \(\mathrm { C } _ { \mathrm { n } }\)
OCR MEI Further Extra Pure 2022 June Q2
12 marks Standard +0.3
2 The matrix \(\mathbf { A }\) is given by \(\mathbf { A } = \left( \begin{array} { r r r } 10 & 12 & - 8 \\ - 1 & 2 & 4 \\ 3 & 6 & 2 \end{array} \right)\).
  1. In this question you must show detailed reasoning. Show that the characteristic equation of \(\mathbf { A }\) is \(- \lambda ^ { 3 } + 14 \lambda ^ { 2 } - 56 \lambda + 64 = 0\).
  2. Use the Cayley-Hamilton theorem to determine \(\mathbf { A } ^ { - 1 }\). A matrix \(\mathbf { E }\) and a diagonal matrix \(\mathbf { D }\) are such that \(\mathbf { A } = \mathbf { E D E } ^ { - 1 }\). The elements in the diagonal of \(\mathbf { D }\) increase from top left to bottom right.
  3. Determine the matrix \(\mathbf { D }\).
OCR MEI Further Extra Pure 2022 June Q3
9 marks Challenging +1.8
3 A sequence is defined by the recurrence relation \(5 t _ { n + 1 } - 4 t _ { n } = 3 n ^ { 2 } + 28 n + 6\), for \(n \geqslant 0\), with \(t _ { 0 } = 7\).
  1. Find an expression for \(t _ { n }\) in terms of \(n\). Another sequence is defined by \(\mathrm { v } _ { \mathrm { n } } = \frac { \mathrm { t } _ { \mathrm { n } } } { \mathrm { n } ^ { \mathrm { m } } }\), for \(n \geqslant 1\), where \(m\) is a constant.
  2. In each of the following cases determine \(\lim _ { n \rightarrow \infty } \mathrm {~V} _ { n }\), if it exists, or show that the sequence is divergent.
    1. \(m = 3\)
    2. \(m = 2\)
    3. \(m = 1\)
OCR MEI Further Extra Pure 2022 June Q4
16 marks Standard +0.8
4 A binary operation, ○, is defined on a set of numbers, \(A\), in the following way. \(a \circ b = \mathrm { k } _ { 1 } \mathrm { a } - \mathrm { k } _ { 2 } \mathrm {~b} + \mathrm { k } _ { 3 }\), for \(a , b \in A\),
where \(k _ { 1 } , k _ { 2 }\) and \(k _ { 3 }\) are constants (which are not necessarily in \(A\) ) and the operations addition, subtraction and multiplication of numbers have their usual notation and meaning. You are initially given the following information about ○ and \(A\).
  • \(A = \mathbb { R }\)
  • \(0 \circ 0 = 2\)
  • An identity element, \(e\), exists for ∘ in \(A\)
OCR MEI Further Extra Pure 2022 June Q5
16 marks Challenging +1.8
5 A surface \(S\) is defined by \(z = f ( x , y )\), where \(f ( x , y ) = y e ^ { - \left( x ^ { 2 } + 2 x + 2 \right) y }\).
    1. Find \(\frac { \partial f } { \partial x }\).
    2. Show that \(\frac { \partial f } { \partial y } = - \left( x ^ { 2 } y + 2 x y + 2 y - 1 \right) e ^ { - \left( x ^ { 2 } + 2 x + 2 \right) y }\).
    3. Determine the coordinates of any stationary points on \(S\). Fig. 5.1 shows the graph of \(z = e ^ { - x ^ { 2 } }\) and Fig. 5.2 shows the contour of \(S\) defined by \(z = 0.25\). \begin{figure}[h]
      \includegraphics[alt={},max width=\textwidth]{76f3559a-f3b3-4a21-878f-adb261dd1236-5_478_686_822_244} \captionsetup{labelformat=empty} \caption{Fig. 5.1}
      \end{figure} \begin{figure}[h]
      \includegraphics[alt={},max width=\textwidth]{76f3559a-f3b3-4a21-878f-adb261dd1236-5_478_437_822_1105} \captionsetup{labelformat=empty} \caption{Fig. 5.2}
      \end{figure}
  2. Specify a sequence of transformations which transforms the graph of \(\mathrm { z } = \mathrm { e } ^ { - \mathrm { x } ^ { 2 } }\) onto the graph of the section defined by \(z = f ( x , 1 )\).
  3. Hence, or otherwise, sketch the section defined by \(z = f ( x , 1 )\).
  4. Using Fig. 5.2 and your answer to part (c), classify any stationary points on \(S\), justifying your answer. You are given that \(P\) is a point on \(S\) where \(z = 0\).
  5. Find, in vector form, the equation of the tangent plane to \(S\) at \(P\). The tangent plane found in part (e) intersects \(S\) in a straight line, \(L\).
  6. Write down, in vector form, the equation of \(L\).
OCR MEI Further Extra Pure 2023 June Q1
7 marks Standard +0.3
1 A surface is defined in 3-D by \(z = 3 x ^ { 3 } + 6 x y + y ^ { 2 }\).
Determine the coordinates of any stationary points on the surface.
OCR MEI Further Extra Pure 2023 June Q2
15 marks Challenging +1.2
2 A sequence is defined by the recurrence relation \(4 \mathrm { t } _ { \mathrm { n } + 1 } - \mathrm { t } _ { \mathrm { n } } = 15 \mathrm { n } + 17\) for \(\mathrm { n } \geqslant 1\), with \(t _ { 1 } = 2\).
  1. Solve the recurrence relation to find the particular solution for \(\mathrm { t } _ { \mathrm { n } }\). Another sequence is defined by the recurrence relation \(( n + 1 ) u _ { n + 1 } - u _ { n } ^ { 2 } = 2 n - \frac { 1 } { n ^ { 2 } }\) for \(n \geqslant 1\), with \(u _ { 1 } = 2\).
    1. Explain why the recurrence relation for \(\mathrm { u } _ { \mathrm { n } }\) cannot be solved using standard techniques for non-homogeneous first order recurrence relations.
    2. Verify that the particular solution to this recurrence relation is given by \(u _ { n } = a n + \frac { b } { n }\) where \(a\) and \(b\) are constants whose values are to be determined. A third sequence is defined by \(\mathrm { v } _ { \mathrm { n } } = \frac { \mathrm { t } _ { \mathrm { n } } } { \mathrm { u } _ { \mathrm { n } } }\) for \(n \geqslant 1\).
  3. Determine \(\lim _ { n \rightarrow \infty } \mathrm { v } _ { \mathrm { n } }\).
OCR MEI Further Extra Pure 2023 June Q3
8 marks Challenging +1.8
3 A surface, \(S\), is defined by \(g ( x , y , z ) = 0\) where \(g ( x , y , z ) = 2 x ^ { 3 } - x ^ { 2 } y + 2 x y ^ { 2 } + 27 z\). The normal to \(S\) at the point \(\left( 1,1 , - \frac { 1 } { 9 } \right)\) and the tangent plane to \(S\) at the point \(( 3,3 , - 3 )\) intersect at \(P\). Determine the position vector of P .
OCR MEI Further Extra Pure 2023 June Q4
15 marks Challenging +1.2
4 The set \(G\) is given by \(G = \{ \mathbf { M } : \mathbf { M }\) is a real \(2 \times 2\) matrix and det \(\mathbf { M } = 1 \}\).
  1. Show that \(G\) forms a group under matrix multiplication, × . You may assume that matrix multiplication is associative.
  2. The matrix \(\mathbf { A } _ { n }\) is defined by \(\mathbf { A } _ { n } = \left( \begin{array} { l l } 1 & 0 \\ n & 1 \end{array} \right)\) for any integer \(n\). The set \(S\) is defined by \(\mathrm { S } = \left\{ \mathrm { A } _ { \mathrm { n } } : \mathrm { n } \in \mathbb { Z } , \mathrm { n } \geqslant 0 \right\}\).
    1. Determine whether \(S\) is closed under × .
    2. Determine whether \(S\) is a subgroup of ( \(G , \times\) ).
    1. Find a subgroup of ( \(G , \times\) ) of order 2 .
    2. By considering the inverse of the non-identity element in any such subgroup, or otherwise, show that this is the only subgroup of ( \(G , \times\) ) of order 2. The set of all real \(2 \times 2\) matrices is denoted by \(H\).
  4. With the help of an example, explain why ( \(H , \times\) ) is not a group.